Chapter 1 · First-Order ODEs

Bernoulli Differential Equations

Transforming Nonlinear to Linear

Dr. Mohamed Mabrok · Qatar University

What is a Bernoulli Equation?

A Bernoulli equation looks linear, but has a nonlinear term $y^n$. A clever substitution will transform it into a linear equation!

Standard Form

      A Bernoulli equation has the form:
    
      $$\frac{dy}{dx} + P(x)y = Q(x)y^n \quad \text{where } n \neq 0, 1$$
    
      The $y^n$ term makes it nonlinear. When $n=0$ or $n=1$, it's already linear and doesn't need the Bernoulli technique.

The magic: we will transform this nonlinear equation into a linear equation using a substitution!

The Magic Substitution

To transform a Bernoulli equation into a linear one, we use the substitution:

v = y^{1-n}

Then differentiate both sides with respect to $x$:

\frac{dv}{dx} = (1-n)y^{-n} \frac{dy}{dx}

From the original equation, divide by $y^n$:

y^{-n}\frac{dy}{dx} + P(x)y^{1-n} = Q(x)

Multiply by $(1-n)$ and substitute $v = y^{1-n}$:

\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)

This is now a first-order LINEAR equation in $v$!

Special Cases: Different Values of $n$

The substitution $v = y^{1-n}$ adapts to each value of $n$:

Value of $n$	Substitution $v = y^{1-n}$	Example Form
$n = 2$	$v = y^{-1}$ (reciprocal)	$\dfrac{dv}{dx} - v = -1$
$n = 3$	$v = y^{-2}$	$\dfrac{dv}{dx} - 2P(x)v = -2Q(x)$
$n = -1$	$v = y^{2}$	$\dfrac{dv}{dx} + 2P(x)v = 2Q(x)$
$n = \frac{1}{2}$	$v = y^{1/2}$ (square root)	$\dfrac{dv}{dx} + \frac{1}{2}P(x)v = \frac{1}{2}Q(x)$

Each case yields a linear equation in the new variable $v$.

The 4-Step Solution Method

STEP 1
        Identify $P(x)$, $Q(x)$, and $n$

        Make sure to normalize first! Divide so the coefficient of $\dfrac{dy}{dx}$ is 1.
      

STEP 2
        Substitute $v = y^{1-n}$

        Transform to: $\dfrac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$
      

STEP 3
        Solve the linear equation

        Use integrating factor $\mu = e^{\int (1-n)P(x)\,dx}$ to solve for $v$.
      

STEP 4
        Back-substitute to find $y$

        Since $v = y^{1-n}$, we have $y = v^{\frac{1}{1-n}}$. Also report $y=0$ as a singular solution.
      

Worked Example 1

Solve: $\dfrac{dy}{dx} + y = y^2$

STEP 1 Identify $P(x)$, $Q(x)$, $n$

P(x) = 1, \quad Q(x) = 1, \quad n = 2

STEP 2 Substitute $v = y^{-1}$ (since $1-n = 1-2 = -1$)

\frac{dv}{dx} - v = -1

STEP 3 Solve: $\mu = e^{-x}$, so $v = 1 + Ce^x$

STEP 4 Back-substitute: $y = \dfrac{1}{v}$

y = \frac{1}{1 + Ce^x}, \quad \text{plus singular solution } y = 0

Worked Example 2

Solve: $xy' + y = x^2 y^2$, $y(1) = 1$

STEP 1 Normalize by dividing by $x$: $y' + \dfrac{y}{x} = xy^2$

P(x) = \frac{1}{x}, \quad Q(x) = x, \quad n = 2

STEP 2 Substitute $v = y^{-1}$

\frac{dv}{dx} - \frac{v}{x} = -x

STEP 3 $\mu = \dfrac{1}{x}$, solve to get $v = -x^2 + Cx$

IC At $x=1$, $y=1$: $v(1) = 1 = -1 + C \Rightarrow C = 2$

y = \frac{1}{x(2-x)}

Worked Example 3

Solve: $\dfrac{dy}{dx} - \dfrac{2y}{x} = -x^2 y^3$

ID $P(x) = -\dfrac{2}{x}$, $Q(x) = -x^2$, $n = 3$

SUB $v = y^{-2}$ (since $1-n = -2$)

\frac{dv}{dx} + \frac{4v}{x} = 2x^2

SOLVE $\mu = x^4$

v = \frac{2x^3}{7} + \frac{C}{x^4}

BACK $y = v^{-1/2}$ since $\dfrac{1}{1-n} = \dfrac{1}{-2} = -\frac{1}{2}$

y = \left(\frac{2x^3}{7} + \frac{C}{x^4}\right)^{-1/2}

Common Pitfalls to Avoid

Mistake 1: Forgetting to Normalize

        The equation must be in the form $\dfrac{dy}{dx} + P(x)y = Q(x)y^n$. Divide by the coefficient of $\dfrac{dy}{dx}$ first!

Mistake 2: Using Wrong Substitution

        The substitution is $v = y^{1-n}$, NOT $y^n$. For $n=2$, use $v = y^{-1}$, not $v = y^2$.

Mistake 3: Forgetting to Scale the Equation

        After substituting $v = y^{1-n}$, both the coefficient of $v$ AND the RHS get multiplied by $(1-n)$.

Mistake 4: Missing the Singular Solution

        Always verify that $y = 0$ is a solution to the original equation. It is a singular solution for all Bernoulli equations.

Mistake 5: Not Back-Substituting

        After solving for $v$, always convert back to $y$ using $y = v^{\frac{1}{1-n}}$. Don't forget this final step!

Key Takeaways

🔑

        Key: $v = y^{1-n}$

        This substitution is the magic that transforms the nonlinear Bernoulli equation into a linear one in $v$.

✨

        Solve the linear equation

        Once you have $\dfrac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$, use the integrating factor method.

♾️

        Always report $y=0$

        The singular solution $y=0$ is always a solution and must be included in your final answer.

💡

        Applications everywhere

        Population dynamics, fluid flow, control systems, and many physical phenomena use Bernoulli equations.

\text{Bernoulli} \;\Longrightarrow\; \text{Substitute } v=y^{1-n} \;\Longrightarrow\; \text{Linear Eq.} \;\Longrightarrow\; \text{Integrate} \;\Longrightarrow\; \text{Back-substitute}